find the value of k for which one root of the equation kx²-14+8 is six times the othera
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The given quadratic equation⇒kx²-14+8=0
Let the roots of the equation be a and b
a=6b
a+b=-(-14)/k (-coeff. of x/coeff. of x²= sum of the roots)
7b=14/k
b=2/k⇒1
ab=8/k (constant/coeff. of x²= product of roots)
6b² =8/k
6(2/k)² =8/k
24/k² =8/k
k=3
Therefore the value of k is 3
Hope it helped you!! :)
Let the roots of the equation be a and b
a=6b
a+b=-(-14)/k (-coeff. of x/coeff. of x²= sum of the roots)
7b=14/k
b=2/k⇒1
ab=8/k (constant/coeff. of x²= product of roots)
6b² =8/k
6(2/k)² =8/k
24/k² =8/k
k=3
Therefore the value of k is 3
Hope it helped you!! :)
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